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$$\int_{-1}^{1}{\int_{x^2}^{1}{cx^2y \ dy \ dx}}$$ Given that: $\int_{-1}^{1}{\int_{x^2}^{1}{cx^2y \ dy \ dx}}=1$, I have to find $c$. I am a bit disoriented by that $x^2$ as the limit of the first integral. I've done so far only integrals with "clear" limits. A bit of help would be useful. Thank you :)

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The inner integral is with respect to $y$, so you hold $x$ fixed and integrate the inner integral first with respect to $y$. That the limit of integration on the inner integral is $x^2$ is fine; $x$ is fixed here, so $x^2$ is just a constant as far as the inner integral is concerned: $$\eqalign{ \int_{-1}^1\int_{x^2}^1 cx^2 y\,dy\,dx &=\int_{-1}^1\biggl[\int_{y=x^2}^{y=1} cx^2 y\,dy\biggr]\,dx\cr &=\int_{-1}^1 \biggl[\ {cx^2y^2\over2}\biggl|_{y=x^2}^{y=1}\,\biggr]\, dx\cr &=\int_{-1}^1 \biggl[ {cx^2\cdot 1^2\over 2}- {cx^2(x^2)^2\over2} \biggr]\, dx. } $$

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solved, thanks :) –  Andrew Apr 18 '12 at 19:54
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The bounds on the inner integral are for values of $y$, so you have $$\begin{align} \int_{-1}^{1} \int_{x^2}^1 cx^2y \;dy \;dx &= c\int_{-1}^1 \left[x^2\frac{1}{2}y^2\bigg\lvert_{x^2}^{1}\right] dx\\ &= \frac{c}{2} \int_{-1}^{1} x^2(1 - x^4)\; dx \\ &= ... \end{align} $$

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many thanks, it helped –  Andrew Apr 18 '12 at 19:55
    
@Andrew No problem. I am glad it helped. –  Thomas Apr 18 '12 at 19:55
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